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I am given a wire with charge density $\lambda$ that is parallel to the $z$ axis and intersects the $xy$ plane at $(d,0,0)$. Space is filled with two different dielectric materials such that in the region $x>0$ the dielectric constant is $\epsilon_1$ and in the region $x<0$ the dielectric constant is $\epsilon_2$. I am given a hint to use the method of images in order to find the electric field in all of space.

I am used to problems involving conductors where the boundary condition is $\phi_S=\phi_0$. Now I guess that the boundary conditions for this problem are $E_1^\parallel=E_2^\parallel$ and $D_1^\perp=D_2^\perp$. My problem with this is that if I place an image wire parallel to the $z$ axis that intersects the $xy$ plane at $(-d,0,0)$, the resulting field will be valid only for the region $x>0$, which would not allow me to produce the required boundary conditions. So how can I use these boundary conditions in order to find a suitable image wire for this problem?

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From the perspective of a x>0 observer, there is a line charge in x>0 and a line charge of a different magnitude in x<0. From the perspective of a x<0 observer, there is only 1 line charge that's in x>0 which has a magnitude different from its observed value in x>0. There are values for all the magnitudes which meet the boundary conditions. If you'd like me to tell you what the values are, I can.

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  • $\begingroup$ My course just published the answers to this and I've seen that their answer is just as you say. However, I don't understand why would the field take such a form? Also, I saw that in the expression for the field they left the field of the original wire with $\epsilon_1$ instead of $\epsilon_0$, whereas in the expression for the field of the image wires there was $\epsilon_0$ (in both cases, x>0 and x<0). What is the reason for that? $\endgroup$
    – Ofek Aman
    Commented Aug 20, 2020 at 7:29
  • $\begingroup$ I don't know why they'd ever use $\epsilon_0$ with how you described the problem. The image charge seen from x>0 should have charge density $\lambda\cdot\frac{\epsilon_1-\epsilon_2}{\epsilon_1+\epsilon_2}$ whereas the other image charge should have charge density $\lambda\cdot\frac{2\epsilon_2}{\epsilon_1+\epsilon_2}$. There are three possible explanations, $\epsilon_2=\epsilon_0$, $\epsilon_2=\epsilon_0\epsilon_r$, or they made a mistake. $\endgroup$
    – Laff70
    Commented Aug 20, 2020 at 14:23
  • $\begingroup$ Hmmm, they indeed got a somewhat different result... Can you explain how you obtained these charge densities? Also, after finding the charge density, what dielectric coefficient will appear in the expression for the fields originating from the respective wires? Will it be the respective dielectric constants of the regions where you put the image wires? $\endgroup$
    – Ofek Aman
    Commented Aug 20, 2020 at 15:01
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    $\begingroup$ I hope this site[ photonics101.com/dielectrics/… ] provides a good explanation. For an observer in x>0, $k=\frac{1}{4\pi\epsilon_1}$. For and observer in x<0, $k=\frac{1}{4\pi\epsilon_2}$. Use the same k for all observed charge. $\endgroup$
    – Laff70
    Commented Aug 20, 2020 at 15:50

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